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Andru [333]
3 years ago
12

tech a says that a tire with more wear on the center of the tread is caused by under inflation of the size tech b says featherin

g of the tire tread is most commonly a result of a excessive toe in or toe out who is correct answer
Engineering
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer: Technician B only is correct.

Explanation:

Tire tread wear on the edges of a tire will typically indicate inflation pressures are lower than specified.

Under inflation occurs when there is more wear on the edges of the tire. A tire is said to be under-inflated whe the contact patch grows and outside edges of the patch takes on the load.

Tire feathering or scuffing is the indicator of excessive positive or negative toe angle that can be detected by stroking your fingertips over the edge of each tread block. A feather edge on the inside of the tread bar shows excess toe-in, whereas a feather edge on the outside of the tread bar shows toe-out. Changes in camber and caster angle affects the toe angle. Changes in suspension height can also affect the toe angle geometry.

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point B where Q_B = 101.25 \ in^3  has the largest Q value at section a–a

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The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the  point that  has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; Q_A = Area \times y_1

where ;

y_1 = (6 - \dfrac{1.5}{2})

y_1 = (6- 0.75)

y_1 = 5.25 \  in

Q_A =(L \times B)  \times y_1

Q_A =(6 \times 1.5)  \times 5.25

Q_A =47.25 \ in^3

For point B ;

Let Q be the moment of the Area B;

SO ; Q_B = Area \times y_2

where ;

y_2 = (6 - \dfrac{1.5 \times 3}{2})

y_2= (6 - \dfrac{4.5}{2}})

y_2 = (6 -2.25})

y_2 = 3.75 \ in

Q_B =(L \times B)  \times y_1

Q_B=(6 \times 4.5)  \times 3.75

Q_B = 101.25 \ in^3

For point C ;

Let Q be the moment of the Area C;

SO ; Q_C = Area \times y_3

where ;

y_3 = (6 - \dfrac{1.5 \times 2}{2})

y_3 = (6 - 1.5})

y_3= 4.5 \  in

Q_C =(L \times B)  \times y_1

Q_C =(6 \times 3)  \times 4.5

Q_C=81 \ in^3

For point D ;

Let Q be the moment of the Area D;

SO ; Q_D = Area \times y_4

since there is no area about point D

Area = 0

Q_D =0 \times y_4

Q_D = 0

Thus; from the foregoing ; point B where Q_B = 101.25 \ in^3  has the largest Q value at section a–a

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